The study of chemical change is the heart of chemistry.

3. The study of chemical change is the heart of chemistry.

4. Law of Conservation of Mass “We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.” --Antoine Lavoisier, 1789 “Atoms are neither created nor destroyed during any chemical reaction.”

5. 3.1 Chemical Equations • Consider a simple chemical equation: 2 H2 + O2 → 2 H2O + : “reacts with”, → : “produces” • reactants and products • coefficients • balanced

6. 3.1 Chemical Equations • Balancing Equations • The difference between a subscript and a coefficient • Subscripts should not be change when balancing equation

7. 3.1 Chemical Equations • Balancing Equations • Consider a chemical reaction:

8. 3.1 Chemical Equations • Balancing Equations • Express the chemical reaction • Balance carbon and hydrogen • Complete the chemical equation by balancing oxygen

9. 3.1 Chemical Equations

10. 3.1 Chemical Equations • Indicating the states of • reactants and products

11. 3.2 Some Simple Patterns of Chemical Reactivity • Combination and decomposition reactions

12. 3.2 Some Simple Patterns of Chemical Reactivity • Combination and decomposition reactions • Figure 3.7

13. 3.2 Some Simple Patterns of Chemical Reactivity

14. 3.2 Some Simple Patterns of Chemical Reactivity • Combustion in air • When hydrocarbons are combusted in air, they react with O2 to form CO2 and H2O. • Combustion of hydrocarbon derivatives • CH3OH, C2H5OH • Glucose (C6H12O6) – oxidation reaction

15. 3.3 Formula Weights • How do we relate the numbers of atoms or molecules to the amounts we measure in the laboratory? • Although we can not directly count atoms or molecules, we can indirectly determine their numbers if we know their masses.

16. 3.3 Formula Weights • Formula and Molecular Weights

17. 3.3 Formula Weights • Percentage composition from formulas • Needed for identifying unknown samples

18. 3.4 Avogadro’s Number & the Mole • The definition of a mole • The amount of matter that contains as many objects (atoms, molecules, etc) as the number of atoms in exactly 12 g of isotopically pure 12C. • Avogadro’s number • The number of objects in 1 mole of matter. • 6.0221421 × 1023

19. 3.4 Avogadro’s Number & the Mole

20. 3.4 Avogadro’s Number & the Mole • Molar Mass • The mass of a single atom of an element (in amu) is numerically equal to the mass (in grams) of 1mol of that element.

21. 3.4 Avogadro’s Number & the Mole • Molar Mass • The molar mass of a substance is the mass in grams of one mole of the substance. • Figure 3.10

22. 3.4 Avogadro’s Number & the Mole 180.0 g/mol 0.02989 mol 84.0 g/mol 6.05 mol

23. 3.4 Avogadro’s Number & the Mole 164.1 g/mol 71.1 g 84.0 g/mol 98.1 g/mol (a) 532 g, (b) 0.0029 g

24. 3.4 Avogadro’s Number & the Mole Molecules C6H12O6

25. 3.5 Empirical Formulas from Analyses • The ratio of the number of moles of each element in a compound gives the subscript in a compound’s empirical formula. • Consider a compound containing Hg & Cl (MWHg 200.6, MWCl 35.5) (73.9% Hg, 26.1% Cl by mass) How to get the empirical formula for the compound? HgCl2

26. 3.5 Empirical Formulas from Analyses • Procedure for calculating an empirical formula from percentage composition. 40.92 g C, 4.58 g H, and 54.50 g O. C:H:O = 3(1:1.33:1) = 3:4:3 C3H4O3

27. 3.5 Empirical formulas from analyses • Molecular formula from empirical formula • The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in its empirical formula. The formula weight of the empirical formula C3H4is 3(12.0 amu) + 4(1.0 amu) = 40.0 amu the molecular formula: C9H12

28. 3.5 Empirical formulas from analyses • Combustion Analysis • A common technique used for the determination of the empirical formula for compounds containing principally carbon and hydrogen. Fig 3.14 Apparatus for combustion analysis.

29. 3.5 Empirical formulas from analyses • Combustion Analysis Mass of O = mass of sample - (mass of C + mass of H) = 0.255 g - (0.153 g + 0.0343 g) = 0.068 g O C3H8O

30. 3.6 Quantitative Information from Balanced Equations

31. 3.6 Quantitative Information from Balanced Equations

32. 3.6 Quantitative Information from Balanced Equations • Procedure for calculating amounts of reactants consumed or products formed in a reaction

33. KClO3 122.55, KCl 74.5, O2 32.00 Answer: 1.76 g

34. MWLiOH 23.95 • 2 LiOH(s) + CO2(g)→Li2CO3(s) + H2O(l) Answer: 3.64 g

36. 3.7 Limiting Reactants Let’s consider a sandwich-making process: If you have 10 slices of bread and 7 slices of cheese, • You will have 5 sandwiches and 2 slices of cheese leftover. • In this case, • Limiting reactant (limiting reagent) : Bd • Excess reactant (excess reagent) : Ch

37. 3.7 Limiting Reactants

38. Sample Exercise 3.18 Calculating the Amount of Product Formed from a Limiting Reactant The most important commercial process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3): N2(g) + 3 H2(g) →2 NH3(g) How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2?

39. Zn: 65.39 g/mol AgNO3: 169.87 g/mol Ag: 107.9 g/mol Zn(NO3)2: 189.39 g/mol • Answer: (a) AgNO3, (b) 1.59 g, (c) 1.39 g, (d) 1.52 g Zn

40. 3.7 Limiting Reactants • Theoretical Yield • The quantity of product that is calculated to form when all of the limiting reactant reacts 146.14 g/mol 84.16 g/mol (a) The theoretical yield is

41. 3.7 Limiting Reactants • Theoretical Yield • The quantity of product that is calculated to form when all of the limiting reactant reacts 146.14 g/mol 84.16 g/mol

42. Exercises

43. Exercises

44. Exercises 3.94 3.94

45. Exercises 3.96 Atomic number 57, Lanthanum

46. Exercises 3.101 3.101